The $n\text{th}$ partial sum of the series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ is given by ${{S}_{n}}=\frac{3n}{3n+1}$. Write a rule for ${{a}_{n}}$.
Answer: An individual term of a series is the difference between two successive partial sums. In particular, ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}\,$. $\begin{aligned} {{S}_{n}}-{{S}_{n-1}}&=\dfrac{3n}{3n+1}-\dfrac{3(n-1)}{3(n-1)+1} \\\\ & ~=\frac{3n}{3n+1}-\frac{3n-3}{3n-2} \\\\ & ~=\frac{3n(3n-2)-(3n-3)(3n+1)}{(3n+1)(3n-2)} \\\\ & ~=\frac{3}{(3n+1)(3n-2)} \end{aligned}$